【LG3320】[SDOI2015]寻宝游戏

题面

洛谷

题解

不需要建虚树的虚树2333。。。

贪心地想一下，起始节点肯定是在关键点上，访问顺序就是\(dfs\)序。

那么对于每次询问，

\[Ans=dis(S_1,S_s)+\sum_{i=1}^{s-1}dis(S_i,S_{i+1})
\]

用\(set\)维护一下就好了

代码

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <set>

using namespace std;

inline int gi() {

    register int data = 0, w = 1;

    register char ch = 0;

    while (!isdigit(ch) && ch != '-') ch = getchar();

    if (ch == '-') w = -1, ch = getchar();

    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();

    return w * data;

}

typedef long long ll;

const int MAX_N = 1e5 + 5;

struct Graph { int to, next, cost; } e[MAX_N << 1];

int fir[MAX_N], e_cnt;

void clearGraph() { memset(fir, -1, sizeof(fir)); }

void Add_Edge(int u, int v, int w) { e[e_cnt] = (Graph){v, fir[u], w}; fir[u] = e_cnt++; }

int N, M;

ll dis[MAX_N], ans;

int dep[MAX_N], size[MAX_N], top[MAX_N], son[MAX_N], fa[MAX_N], dfn[MAX_N], tim;

void dfs1(int x) {

	size[x] = 1, dep[x] = dep[fa[x]] + 1;

	for (int i = fir[x]; ~i; i = e[i].next) {

		int v = e[i].to; if (v == fa[x]) continue;

		dis[v] = dis[x] + e[i].cost;

		fa[v] = x, dfs1(v), size[x] += size[v];

		if (size[v] > size[son[x]]) son[x] = v;

	}

}

void dfs2(int x, int tp) {

	top[x] = tp, dfn[x] = ++tim;

	if (son[x]) dfs2(son[x], tp);

	for (int i = fir[x]; ~i; i = e[i].next) {

		int v = e[i].to; if (v == fa[x] || v == son[x]) continue;

	    dfs2(v, v);

	}

}

int LCA(int u, int v) {

	while (top[u] != top[v]) {

		if (dep[top[u]] < dep[top[v]]) swap(u, v);

		u = fa[top[u]];

	}

	return dep[u] < dep[v] ? u : v;

}

ll DIS(int x, int y) { return dis[x] + dis[y] - 2ll * dis[LCA(x, y)]; }

struct cmp { bool operator () (int x, int y) const { return dfn[x] < dfn[y]; } } ;

set<int, cmp> s = set<int, cmp>(cmp());

set<int, cmp> :: iterator ite;

bool vis[MAX_N];

int main () {

#ifndef ONLINE_JUDGE

    freopen("cpp.in", "r", stdin);

#endif

	clearGraph();

	N = gi(), M = gi();

	for (int i = 1; i < N; i++) {

		int u = gi(), v = gi(), w = gi();

		Add_Edge(u, v, w), Add_Edge(v, u, w);

	}

	dfs1(1), dfs2(1, 1);

	while (M--) {

		int x = gi();

		if (vis[x]) {

			ite = s.find(x);

			if (s.size() < 3) { ans = 0; goto P1; }

			int pre, nxt;

			if (ite == s.begin()) pre = *--s.end();

			else pre = *(--ite), ++ite;

			if (ite == --s.end()) nxt = *s.begin();

			else nxt = *(++ite), --ite;

			ans -= DIS(pre, x) + DIS(x, nxt) - DIS(pre, nxt);

		  P1 :

			s.erase(ite);

		} else {

			ite = s.insert(x).first;

			if (s.size() == 1) goto P2;

			int pre, nxt;

			if (ite == s.begin()) pre = *--s.end();

			else pre = *(--ite), ++ite;

			if (ite == --s.end()) nxt = *s.begin();

			else nxt = *(++ite), --ite;

			ans += DIS(pre, x) + DIS(x, nxt) - DIS(pre, nxt);

		  P2: ;

		}

		vis[x] ^= 1;

		printf("%lld\n", ans);

	}

	return 0;

}

巴特西

【LG3320】[SDOI2015]寻宝游戏

【LG3320】[SDOI2015]寻宝游戏

题面

题解

最新文章

热门文章